From Discrete Math Rosen textbook 8th edition Section 1.4 Exercises:
Exercise 48-51 establish rules for null quantification that we can use when a quantified variable does not appear in part of a statement.
Establish these logical equivalences, where x does not occur as a free variable in A. Assume that the domain is nonempty.
Exercise 48:
a) (∀xP(x)) ∨ A ≡ ∀x(P(x) ∨ A)
b) (∃xP(x)) ∨ A ≡ ∃x(P(x) ∨ A)
Exercise 49:
a) (∀xP(x)) ∧ A ≡ ∀x(P(x) ∧ A)
b) (∃xP(x)) ∧ A ≡ ∃x(P(x) ∧ A)
Exercise 50:
a) ∀x(A → P(x)) ≡ A → ∀xP(x)
b) ∃x(A → P(x)) ≡ A → ∃xP(x)
Exercise 51:
a) ∀x(P(x) → A) ≡ ∃xP(x) → A
b) ∃x(P(x) → A) ≡ ∀xP(x)→ A
QUESTION 1: Why do they have to mention that "...where x does not occur as a free variable in A." here? I feel it can be for 2 reasons:
So many of the expressions can actually be propositions since if x were a free variable in A, then expressions like (∀xP(x)) ∨ A wouldn't be considered propositions since A's x is not quantified nor assigned a value, the requirements for A to be a proposition and hence make this conjunction a proposition
Mainly relating to definition of null quantification, so A is not affected by any quantifier of A, hence we're able to pull it out like this and make logical equivalences.
This will also help clarify another part of the textbook:
Express the statement “If a person is female and is a parent, then this person is someone’s mother” as a logical expression involving predicates, quantifiers with a domain consisting of all people, and logical connectives.
Solution: The statement “If a person is female and is a parent, then this person is someone’s mother” can be expressed as “For every person x, if person x is female and person x is a parent, then there exists a person y such that person x is the mother of person y.” We introduce the propositional functions F(x) to represent “x is female,” P(x) to represent “x is a parent,” and M(x, y) to represent “x is the mother of y.” The original statement can be represented as
∀x((F(x) ∧ P(x)) → ∃yM(x, y)).
Using the null quantification rule in part (b) of Exercise 49 in Section 1.4, we can move ∃y to the left so that it appears just after ∀x, because y does not appear in F(x) ∧ P(x). We obtain the logically equivalent expression
∀x∃y((F(x) ∧ P(x)) → M(x, y)).
Where I guess they meant to say Exercise 50 part b instead, and I guess they just did (using exericse 50b result):
∀x( (F(x) ∧ P(x)) → ∃yM(x, y) )
≡
∀x( ∃y[ (F(x) ∧ P(x)) → M(x, y) ] )
≡
∀x∃y((F(x) ∧ P(x)) → M(x, y))
Where they just focused on the inner statement like I showed here.
QUESTION 2: So this leads me to another question, which is when looking at inner statements, do we assume we're looking at some particular x in the domain, since otherwise,
(F(x) ∧ P(x)) → ∃yM(x, y) wouldn't really be a proposition since the x is quantified nor assigned a value here?
Kindly please let me know if I'm correct and further help clarify my understanding of null quantification.